Chapter 9, Problem 172RQ

a)

To determine

The maximum temperature and pressure during the cycle.

Explanation of Solution

Calculation:

Draw the $P-\nu$ for an ideal Otto cycle as shown in Figure 1.

Calculate the clearance volume of one cylinder.

  $\begin{array}{l}r=\frac{{\text{V}}_c+{\text{V}}_d/4}{{\text{V}}_c}\\ 11=\frac{{\text{V}}_c+\frac{1.8\text{L}}{4}}{{\text{V}}_c}\\ 11=\frac{{\text{V}}_c+\frac{1.8\text{L}\left(\frac{1{\text{m}}^3}{1000\text{L}}\right)}{4}}{{\text{V}}_c}\\ {\text{V}}_c={\text{V}}_2=0.000045{\text{m}}^3\end{array}$

Calculate the volume at state 1.

  $\begin{array}{c}{\text{V}}_1={\text{V}}_c+\frac{{\text{V}}_d}{4}\\ =0.000045{\text{m}}^3+\frac{1.8\text{L}}{4}\\ =0.000045{\text{m}}^3+\frac{1.8\text{L}}{4}\left(\frac{1{\text{m}}^3}{1000\text{L}}\right)\\ =0.000495{\text{m}}^3\end{array}$

Calculate the mass of the air.

  $\begin{array}{c}{P}_1{\text{V}}_1=mR{T}_1\\ m=\frac{P_1{\text{V}}_1}{R{T}_1}\\ =\frac{90\text{kPa}\times 0.000495{\text{m}}^3}{0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\times 50°\text{C}}\\ =\frac{90\text{kPa}\times 0.000495{\text{m}}^3}{0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\times \left(50+273\right)\text{K}}\\ =0.0004805\text{kg}\end{array}$

The properties at during process 1-2 are as follows:

  $\begin{array}{l}{T}_1={50}^{°}\text{C}=323\text{K}\to u_1=230.88\text{kJ}/\text{kg}\\ \begin{array}{l}{T}_1={50}^{°}\text{C}=323\text{K}\hfill \\ P_1=90\text{kPa}\hfill \end{array}\}{s}_1=5.8100\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

  $\begin{array}{l}{v}_1=\frac{{\text{V}}_1}{m}=\frac{0.000495{\text{m}}^3}{0.0004805\text{kg}}=1.0302{\text{m}}^3/\text{kg}\\ {\text{V}}_2={\text{V}}_c=0.000045{\text{m}}^3\\ v_2=\frac{{\text{V}}_2}{m}=\frac{0.000045{\text{m}}^3}{0.0004805\text{kg}}=0.09364{\text{m}}^3/\text{kg}\end{array}$

  $\begin{array}{l}\begin{array}{l}{s}_2=s_1=5.8100\text{kJ}/\text{kg}.\text{K}\hfill \\ v_2=0.09364{\text{m}}^3/\text{kg}\hfill \end{array}\}{T}_2=807.3\text{K}\\ T_2=807.3\text{K}\to u_2=598.33\text{kJ}/\text{kg}\\ P_2=P_1\frac{{\text{V}}_{\text{1}}}{{\text{V}}_{\text{2}}}\frac{T_2}{T_1}\\ P_2=(90\text{kPa})\left(\frac{0.000495{\text{m}}^3}{0.000045{\text{m}}^3}\right)\left(\frac{807.3\text{K}}{323\text{K}}\right)\\ P_2=2474\text{kPa}\end{array}$

During process 2-3:

  $\begin{array}{l}{Q}_{\text{in}}=m\left(u_3-u_2\right)\\ 1.5\text{kJ}=(0.0004805\text{kg})\left(u_3-598.33\right)\text{kJ}/\text{kg}\\ u_3=3719.8\text{kJ}/\text{kg}\end{array}$

  $\begin{array}{l}{u}_2=598.33\text{kJ}/\text{kg}\to T_3=4037\text{K}\\ P_3=P_2\left(\frac{T_3}{T_2}\right)=(2474\text{kPa})\left(\frac{4037\text{K}}{807.3\text{K}}\right)\\ P_3=12,375\text{kPa}\\ \begin{array}{l}{T}_3=4037\text{K}\hfill \\ P_3=12,375\text{kPa}\hfill \end{array}\}{s}_3=7.3218\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Thus, the maximum temperature and pressure during the cycle are $4037\text{K}$ and $\text{12375 kPa}$ respectively.

b)

To determine

The net work per cycle per cylinder and the thermal efficiency of the cycle.

Explanation of Solution

Calculation:

During process 3-4:

  $\begin{array}{l}{s}_4=s_3=7.3218\text{kJ}/\text{kg}.\text{K}\hfill \\ v_4=v_1=1.0302{\text{m}}^3/\text{kg}\hfill \end{array}\}\to \{\begin{array}{l}{T}_4=2028\text{K}\\ u_4=1703.6\text{kJ}/\text{kg}\end{array}$

  $\begin{array}{c}{P}_4=P_3\frac{{\text{V}}_3}{{\text{V}}_4}\frac{T_4}{T_3}=(12,375\text{kPa})\left(\frac{1}{11}\right)\left(\frac{2028\text{K}}{4037\text{K}}\right)\\ =565\text{kPa}\end{array}$

During process 4-1:

  $\begin{array}{c}{Q}_{\text{out}}=m\left(u_4-u_1\right)\\ =(0.0004805\text{kg})(1703.6-230.88)\text{kJ}/\text{kg}\\ =0.7077\text{kJ}\end{array}$

Calculate the net power output $W_{net}$.

  $\begin{array}{c}{W}_{net}=Q_{in}-Q_{out}\\ W_{net}=1.5\text{kJ}-0.7077\text{kJ}\\ =\text{0}\text{.792}\text{kJ}\end{array}$

Thus, the net power output is $\text{0}\text{.792}\text{kJ}$.

Calculate the thermal efficiency of the cycle $\left({\eta }_{th}\right)$.

  $\begin{array}{c}{\eta }_{th}=\frac{W_{\text{net}}}{Q_{\text{in}}}\\ =\frac{0.792}{1.5}\\ =0.528\end{array}$

Thus, the thermal efficiency is $0.528$.

c)

To determine

The mean effective pressure of the cycle.

Explanation of Solution

Calculation:

Calculate the mean effective pressure for an ideal Otto cycle $\left(\text{MEP}\right)$.

  $\begin{array}{c}\text{MEP}=\frac{W_{\text{net}}}{v_1-v_2}\\ =\frac{\text{0}\text{.792}\text{kJ}}{0.000495{\text{m}}^3-0.000045{\text{m}}^3}\\ =1761\text{kPa}\end{array}$

Thus, the mean effective pressure of the cycle is $1761\text{kPa}$.

d)

To determine

The power output for an engine speed of 3000 rpm.

Explanation of Solution

Calculation:

Calculate the power produced by the engine $\left({\dot{W}}_{\text{net}}\right)$.

  $\begin{array}{c}{\dot{W}}_{\text{net}}=\frac{\left(n_{cyl}\times W_{\text{net}}\right)\dot{n}}{n_{\text{rev}}}\\ =\frac{\left(\text{4}\text{cylinder}\times 0.792\text{kJ/cylinder-cycle}\right)\text{3000}\text{rev/min}}{\text{2}\text{rev/cycle}}\\ =\frac{\left(\text{4}\text{cylinder}\times 0.792\text{kJ/cylinder-cycle}\right)\text{3000}\text{rev/min}\left(\frac{1\text{min}}{60\text{s}}\right)}{\text{2}\text{rev/cycle}}\\ =79.2\text{kW}\end{array}$

Thus, the power output for an engine speed of 3000 rpm is $79.2\text{kW}$.